As regards the limit $a\to0$, an integration by parts yields
$$
L(a)=a\int_0^{+\infty}\mathrm e^{-x}\log(1+x/a)\mathrm dx=aN(a)-a\log(a),
$$
with
$$
N(a)=\int_0^{+\infty}\mathrm e^{-x}\log(a+x)\mathrm dx.
$$
Another integration by parts yields
$M(a)=a-aL(2a)$. Since $N(a)=N(0)+o(1)$ and $N(0)$ is finite, one gets
$$
a^{-1}L(a)=-\log(a)+O(1),\quad a^{-1}M(a)=1+O(1).
$$
Since $\mathrm e^{-a}=1+O(a)$ and $\mathrm e^{-2a}=O(1)$,
one sees that $J(a)$ and
$J(a)+K(a)$ are both $-\log(a)+O(1)$. Finally, when $a\to0$,
$I(a)=-\log(a)+O(1)$, and in particular,
$$
\color{red}{\lim\limits_{a\to0}\frac{I(a)}{\log a}=-1}.
$$